Lp estimates for the bilinear Hilbert transform

  1. Michael Lacey and
  2. Christoph Thiele
  1. School of Mathematics, Georgia Institute of Technology, Atlanta, GA 30332; and Department of Mathematics, Universität Kiel, D-24118 Kiel, Germany

Abstract

For the bilinear Hilbert transform given by: Formula we announce the inequality ∥H fgp3K p1 ,p2fp1gp2, provided 2 < p 1, p 2 < ∞, 1/p 3 = 1/p 1 + 1/p 2 and 1 < p 3 < 2.

We announce a partial resolution to long standing conjectures concerning the operator known as the bilinear Hilbert transform, defined as follows: Formula This operation is initially defined only for certain functions f and g, for instance those in the Schwartz class on ℝ. The conjectures concern the extension of H to a bounded operator on L p spaces. We have proved:

Theorem 1. H extends to a bounded operator on Lp1 × Lp2 into Lp3, provided 2 < p1, p2 < ∞ and 1 < p3 < 2, where 1/p3 = 1/p1 + 1/p2.

Some 30 years ago, in connection with the Cauchy integral on Lipschitz curves, Calderón (1) raised the question of H mapping L 2 × L 2 into L 1; this inequality is true. Indeed, the bilinear Hilbert transform maps into L p3  provided only that p 3 > 2/3.

Study of the bilinear Hilbert transform is intimately related to Carleson’s theorem (2) asserting the pointwise convergence of Fourier series. A seminal result, it has received two proofs, with the alternative proof provided by Fefferman (3). These proofs have provided us with ingenious and complementary methods of time frequency analysis. A similar analysis seems necessary to understand H, and so our proof entails significant aspects of both Carleson’s and Fefferman’s proofs. We give a description of our proof, with details presented in their most concrete form. Complete proofs, which appear in ref. 4, require definitions and constructions somewhat more general than those presented here.

The bilinear Hilbert transform must be broken into scales and the frequency behavior of each scale understood. Hence we replace the kernel 1/y with Σj=−∞ 2jρ(2j y), where ρ is a Schwartz function with Fourier transform ρ̂(ξ) = ∫ e −2πixξρ(x) dx supported on [1/2, 2). For each j, consider: Formula which has bilinear symbol ρ̂(2j(ξ − θ)). More specifically, Formula Therefore, if f is supported in frequency on the interval [n2j, (n + 1)2j], then H j fg(x) acts on the inverse Fourier transform of ĝ(ξ)1[(n + 1/2)2j, (n + 3)2j](ξ), and is supported in frequency on the interval [(2n + 1/2)2j, (2n + 4)2j]. The differing rates of translation make these three intervals distinct.

It is important to note that the location of the intervals is arbitrary, and therefore, for all j and j′, the inner product of H j fg and H j fg need not tend to zero as |jj′| tends to infinity. The analysis of H must be done in terms of both time and frequency.

Instead of proceeding with a decomposition of H, we define a model of it adapted to the combinatorics of the time–frequency plane. Let 𝒟 be a dyadic grid in ℝ. Call I × ω ∈ 𝒟 × 𝒟 a tile if |I|·|ω| = 1. The interval ω is a union of four dyadic subintervals of equal length, ω1, ω2, ω3, and ω4, which we list in ascending order. Thus, ξi < ξj for all 1 ≤ i < j ≤ 4 and ξj ∈ ωj. (We will only use ωj for j = 1, 2, 3.) We adopt the notation t = I t × ωt and tj = I t × ωtj for j = 1, 2, 3. Fix a Schwartz function φ with φ̂ supported on [−1/8, 1/8], in addition require that ∫ φ(x − 16n)φ(x) dx = 0 for all integers n. Set for all tiles t and j = 1, 2, 3, Formula where c(J) denotes the center of the interval J.

Then our model of the bilinear Hilbert transform is Formula which is initially defined only for Schwartz functions f 1 and f 2. We emphasize that the sum extends over all tiles, and hence all scales. The analogue of Theorem 1 is

Theorem 2. ℳ extends to a bounded operator on Lp1 × Lp2 into Lp′3, provided 2 < p1, p2 < ∞ and 1 < p′3 = (1/p1 + 1/p2)−1 < 2.

With more liberal notions of “grid,” “tile,” and “φtj,” the bilinear Hilbert transform is in the convex hull of terms like our model ℳ.

In the present situation we can give a proof by way of duality. Thus we take f 3L p3 and show that: Formula The sum is over positive quantities; namely, the decomposition above already captures all of the cancellation necessary for convergence of the sums. It also shows that the sum defining ℳ is unconditionally convergent in t. And as each f jL pj, where p j > 2, it follows that each function is locally square integrable. As it turns out, L 2 arguments are decisive in proving Theorem 2.

We localize the sum above in the x variable by setting: Formula Certainly ∫ F t(x) dx = |I t|−1/2j=1 3 |〈f j, φtj〉|. And so we show that F(x) = Σt F t(x) is integrable. This follows from a weak-type result: for p j as above, there is a δ > 0 so that for all |r jp j| < δ, 1 ≤ j ≤ 3, the operator F(x) maps L r1 × L r2 × L r3 into L r,∞, where 1/r = 1/r 1 + 1/r 2 + 1/r 3. Then a variant of the Marcinkiewicz interpolation theorem due to Janson (5) implies the strong-type inequality.

A single instance of the weak-type inequality is: Formula for some constant K. But this inequality implies the weak-type result, because F commutes with dilations by powers of 2, and so it suffices to establish this last inequality. These observations are useful since some of our estimates begin to break down on exceptional sets of small measure. Due the localization of F t in the time variable and the fact that we only aim for a distributional inequality, we can delete tiles t whose time coordinate falls in a set of bounded measure.

The combinatorics of the time frequency plane enter in by way of the partial order on the tiles given by t < t′ if I tI t and ω ⊃ ω′. Note that t and t′ are not comparable with respect to < if and only if tt′ = ∅. Being disjoint suggests orthogonality for the functions φtj and φtj, the dominant theme of Lemmas 1–3 we state below.

Call a collection of tiles T a Carleson–Fefferman (CF) set with top q if t < q for all tT. Thus ωq ∩ ωt ≠ ∅ for tT. Call T a j-CF set if T is a CF-set for which the intervals ωtj intersect for all tT. Notice that if T is a 1-CF set, say, then the intervals {ωtj|tT} are pairwise disjoint for j = 2, 3. Therefore, by application of Cauchy–Schwartz: Formula Formula Notice that the last two square functions are Littlewood–Paley g functions, albeit conjugated by an exponential to account for the location of the CF set in frequency.

This estimate forms the motivation for Lemma 1 below, which formalizes a decomposition of the set of tiles that is fundamental to our argument.

Lemma 1. Fix pi > 2. There is a δ > 0 and an ɛ0 > 0 and a constant K so that for all |ri − pi| < δ and 0 < ɛ < ɛ0, the following holds. The collection of all tiles S is a union: Formula with these properties. First, S 0 is trivial in that: Formula Then S n,i,j is a union of disjoint i-CF sets T q with tops q ∈ S * n,i,j, and: Formula Here, recall that 1/r = Σi 1/ri, which can be taken arbitrarily close to 1. And, most significantly, for t = min i{pi/2} − ɛ, Formula With Lemma 1 in place, we estimate: Formula Formula The last sum is finite as r is arbitrarily close to one, while t + ɛ = min{p i/2} > 1 is a fixed distance from one. Therefore, with Eq. 3, Eq. 1 holds.

We cannot give the complete construction of the S n,i,j, but rather the initial steps, in which the nearly orthogonal classes of φti are identified. First, we make an important comparison to a maximal function. If T q is an i-CF set with top q, we have for ji, Formula Here M 2 g is the maximal function (M|g|2)1/2. Thus the set F = ∪i{M 2 f i > C −1} has bounded measure and we define S 0 = {s|I sF}, making Eq. 4 trivial. For all i-CF sets T with top q, and TS/S 0, we have Δ(T, j) ≤ 1, for ji.

The remaining construction is inductive. Assume that the S m,i,j are defined for all m < n and all i, j, in such a way that for S r = S/∪m<ni,j S m,i,j) we have: Formula and for any i-CF set T qS r with top q, Δ(T q, j) ≤ 2n/rj +2, for ji. As the same inequality applies to each sub-CF set of T, we conclude that: Formula We define S* n,1,1 to be the set of maximal tiles q with |〈f 1, φq,1〉| ≥ 2n/r1−1 Formula, and take S n,1,1 to consist of all tiles t so that t1 < q for some qS* n,1,1. These tiles are removed, and then S n,i,i is defined similarly for i = 2, 3. After the deletion of the tiles D 0 = ∪i=1 3 S n,i,i, we have |〈f i, φti〉| ≤ 2n/ri −1 Formula for all tiles tS r = S r/D 0.

The set S n,1,2 has a slightly different construction. Consider 1-CF sets T qS r with top q so that Δ(T, 2) ≥ 2n/r2 +1. We take T q to be the maximal 1-CF set with this property. Let q(1) be such a top, which is maximal with respect to <, and in addition sup{ξ|ξ ∈ ωq} is maximal. Remove the tiles T q(1), and repeat this procedure to define T q(2) and so on. S n,1,2 is then ∪ T q(ℓ) and S* n,1,2 = {q(ℓ)|ℓ ≥ 1}. Observe that for any 1-CF set TS r/S n,1,2, we have Δ(T, 2) ≤ 2n/r2 +1. These procedures are repeated inductively to define the S n,i,j for all n, i, j.

With the construction above it is elementary to check that these properties hold. Formula And in the case of ij, the collection S n,i,j is a union of disjoint i-CF sets T q, with qS* n,i,j, for which: Formula These last two bounds differ by a factor of 4, which is relevant below. See the comments concerning the minimal tiles immediately following Lemma 3 below. To achieve Eq. 4, one must delete some tiles t, using Eq. 2, the upper bounds Eqs. 6 and 7, and the control on the number of trees given in Eq. 5.

The essence of the matter lies in the control of the number of CF-sets, which is in the verification of Eq. 5. Eq. 5 relies upon the inequalities in the previous paragraph and Lemma 2 and 3 below, which address the issue of almost orthogonality.

Let us consider S n,1,1, say. The tiles S* n,1,1 are maximal and therefore pairwise disjoint, which suggest weak orthogonality for the collection of functions {φq,1|qS* n,1,1}. If they were in fact orthogonal, Bessel’s inequality and Eq. 8 implies: Formula While f 1 is not in L 2, this inequality can be strengthened to an analogous form for L r for r > 2.

However, disjointness of tiles does not imply orthogonality, because the functions φq,1 are not compactly supported in the x variable. Indeed, by our choice of φ, for two tiles t and s we have 〈φti, φsi〉 = 0 if ωti ∩ ωsi = ∅ or ωti = ωsi. But if ωti ωsi then for all n ≥ 0: Formula If we assume that a stronger separation of the tiles in the x variable, then we would expect orthogonality. And in this direction we have:

Lemma 2. For n ≥ 1 there are constants K and Kn so that the following holds for all A ≥ 1. Let S be any collection of tiles so that: Formula Here, for an interval I, AI denotes the interval with the same center as I and length A|I|. Set NS(x) = Σt∈S 1 It(x). Then: Formula A further combinatorial lemma asserts that if the tiles {ti|tS′} are merely disjoint, then after deleting tiles t for which I t falls in an exceptional set of small measure, S′ is a union of O(A 3) collections of tiles S that satisfy the stronger disjointness condition (Eq. 11).

The previous lemma is essential in obtaining Eq. 5 for the classes S n,i,i. A corresponding lemma is necessary for the S n,i,j, with ij, with Eq. 9 replacing the role of Eq. 8. It is:

Lemma 3. For n ≥ 1 there are constants K and Kn so that the following holds for all A ≥ 1. Let S be a union of j-CF sets T q with tops q ∈ S* . Suppose that: Formula and for t ∈ T q, q ∈ S* and i ≠ j fixed, Formula Set NS(x) = Σq∈S* 1 Iq(x). Then: Formula Notice that in a j-CF set T q, the tiles {ωti|tT q} are pairwise disjoint. Thus Eq. 12 is stronger than merely asserting that the tiles {ωti|tT} are pairwise disjoint. With the construction of the S n,i,j for ij given above, Eq. 12 is true after deleting the minimal tiles S n,i,j min in S n,i,j. The minimal tiles are controlled with the first half of Eq. 9 and the observation that ΣsSn,i,jmin 1 Is(x) ≤ N n,i,j(x) for all x.

The method of proof of both Lemmas 2 and 3 is similar. For instance, in Lemma 2, one considers the operator: Formula If S is finite, this is a compact self-adjoint operator, with maximal eigenvalue B. It suffices to estimate B, as for all fL 2, ΣtS |〈f, φti〉|2 = 〈f, 𝒮S f〉 ≤ Bf2 2. Consider a normalized extremal eigenfunction f of 𝒮S. One then estimates: Formula which is expanded in diagonal and off-diagonal terms. The diagonal term is ΣtS |〈f, φti〉|2 = 〈f, 𝒮S f〉 ≤ B, which is an adequate estimate for B 2. The off-diagonal term is by Cauchy–Schwarz, Formula Formula The innermost sum is bounded by C n A n Formula infxItMMf(x). This is seen by invoking the estimate |〈f, φsi〉| ≤ K Formula infxIs Mf(x), using Eq. 10 and carefully exploiting the geometry of the tiles via the assumption 11.

One then sees that the off-diagonal term is no more than: Formula This, with the diagonal estimate, proves that B 2B + C n B 1/2 A nN S, whence follows Lemma 2.

The research outlined herein is the product of several years of effort (6, 7).

Acknowledgments

The success of our efforts is due to the guidance and encouragement we have received from R. Coifman. M.L. has been supported by the National Science Foundation. M.L. and C.T. acknowledge the support of a North Atlantic Treaty Organization travel grant.

Footnotes

  • Charles Fefferman, Princeton University, Princeton, NJ

  • Abbreviation: CF, Carleson–Fefferman.

References

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